91影视

The given data can be listed below as,

• The mass of the car is,m .
• The radius of the track is,r = 5.00m .
• The normal force at the bottom is,N = 250 mg .

The normal force is the force of contact applied by a surface on an object. The normal force only occurs when the two surfaces are in contact.

The expression for the normal force using newton鈥檚 second law in a circular motion can be expressed as,

$N-mg=\frac{m{v}^2}{r}$

Here N is the normal force, m is the mass of the car, g is the acceleration due to gravity, v is the velocity, and r is the radius of the track.

Substitute 250 mg forN , 9.8 m/s ²for g, 500 m for r in the above equation.

$$\begin{gathered} 2.50mg-mg=\frac{m{v}^2}{5.00\text{m}} \\ mg(2.50-1)=\frac{m{v}^2}{5.00\text{m}} \\ 1.5g脳5.00\text{m}=v^2 \\ v=\sqrt{1.5脳9.8{\text{m/s}}^{\text{2}}脳5.00\text{m}} \\ =8.57\text{m/s} \end{gathered}$$

Hence, the velocity is,8.57 m/s .

The expression for the time period can be expressed as,

$T=\frac{2蟺r}{v}$

Substitute 500 m for r ,8.57 for v , 3.14 for in the above equation.

$$\begin{gathered} T=\frac{2脳3.14脳5.00\text{m}}{8.57\text{m/s}} \\ =3.66sec \end{gathered}$$

Hence, the required time is, 3.66 sec.